From [1]} with \(n=N-j-2\)  and \(m=d-j-3\)  if \(d-j-2>0\)  (the assumption \(2m<n+1\)  is satisfied since \(N-2d+j\ge 0\) ) we obtain
\( \frac{S_{j+1}}{\binom{N-j-2}{d-j-2}} \le \frac{d-j-2}{N-d+1} \cdot \frac{N-d+2}{N-2d+j+5} \le \frac{d-j-2}{N-d+1} \cdot \frac{N-d+2}{N-2d+j+1}.\) 
